Question

solve this..............

Answer 1

$let \\ \sqrt{5 + \sqrt{5 + \sqrt{5...} } } = x \\ then \\ {x}^{2} = 5 + \sqrt{5 + \sqrt{5 + ..} } \\ {x}^{2} = 5 + x \\ {x}^{2} - x - 5 = 0 \\ x = \frac{1 + \sqrt{21} }{2} or \\ x = \frac{1 - \sqrt{21} }{2} \\ x \: cant \: be \: negative \: \\ so \\ x = \frac{1 + \sqrt{21} }{2}$

Answer 2

Answer is in the attachment.

Take the whole thing as x.

Square it. You will finally get a quadratic equation in x.

Solve it to get x.

Check if the solution is valid or not.( one of the solution is negative which is not acceptable in this case. so the other one is the answer)